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Continuous-Time Markov Chains of the Ehrenfest Type with Applications*

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Continuous-Time Markov Chains of the Ehrenfest Type with Applications* 数理解析研究所講究録 1193 巻 2001 年 79-103 79 ASYMPTOTIC BEHAVIOR OF CONTINUOUS-TIME MARKOV CHAINS OF THE EHRENFEST TYPE WITH APPLICATIONS* 神奈川大学 工学部 成田清正 (Kiyomasa Narita) Faculty of Engineering, Kanagawa Univ. Abstract Kac [6] considers the discrete-time Markov chain for the original Ehrenfest urn model and gives the expression of the $n$-step transition probabilities in terms of the Krawtchouk polyno- mials; moreover, Kac’s formulas for transition laws show that the model yields Newton’s law of cooling at the level of the evolution of the averages. Here we first give a version of Newton’s law of cooling for Dette’s [4] Ehrenfest urn model. We next discuss the continuous-time formulation of Markov chain for Krafft and Schaefer’s [9] Ehrenfest urn model and calculate probability law of transitions and first-passage times in both the reduced description with states counting the number of balls in the specified urn and the full description with states enumerating the vertices of the $N$-cube. Our results correspond to a generalization of the work of Bingham [3]. We also treat several applications of urn models to queueing network and reliability theory. 1. Introduction Let $s$ and $t$ be two parameters such that $0<s=<1$ and $0<t=<1$ , and consider the $\{0,1, \ldots , N\}$ and transition probabilities homogeneous Markov chain with state . $(1- \frac{i}{N})s$ , if $j=i+1.$ ’ $\frac{i}{N}t$ , if $j=i-1$ , $p_{ij}=\{$ (1.1) $0<<=^{j}=^{N}$ $1-(1- \frac{i}{N})s-\frac{i}{N}t$ , if $j=i$ , ’ $0$ , otherwise, which is proposed by Krafft and Schaefer [9]. In case $s=t=1$ , the Markov chain is the same as P. and T. Ehrenfest [5] use for illustration of the process of heat exchange between two bodies that are in contact and insulated from the outside. The temperatures are assumed to change in steps of one unit and are represented by the numbers of balls in two urns. The two urns are marked I and II and there are $N$ balls labeled 1, 2, . , $N$ and distributed in two urns. The chain is in state $i$ when there are $i$ balls in I. At each $1/N$ ball point of time $n$ one ball is chosen at random (i.e. with equal probabilities among *This research was supported in part by $\mathrm{G}\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{t}_{-}\mathrm{i}\mathrm{n}_{-\mathrm{A}\mathrm{i}}\mathrm{d}$ for General Scientific Research (No. 10640138), The Ministry of Education, Science, Sports and Culture. 80 numbers 1, 2, . , $N$ ) and moved from its urn to other urn by the following way. If it is in urn I, it is placed in II with probability $t$ and returned to I with probability l-t. If it $\Pi$ is in , it is placed in I with probability $s$ and returned to II with probability l-s. The case where $s+t=1$ is the one-parameter Ehrenfest urn model as investigated by Karlin and $\mathrm{M}\mathrm{c}\mathrm{G}\mathrm{r}\mathrm{e}\mathrm{g}\mathrm{o}\mathrm{r}[7]$ . The case where $s=t$ is the thinning of the Ehrenfest urn model as proposed by Vincze [14] and Tak\'acs [13]. Remark 1.1. The process of the Markov chain in the model (1.1) corresponds to a $N$ ‘reduced description’ with $N+1$ states, $0,1,$ $\cdots,$ counting the number of balls in urn I. If we use a ‘full description’ with $2^{N}$ states, representing the possible configurations by $N$-tuples $i=$ $(i_{1}, \ldots, i_{k}, \ldots , i_{N})$ , where $i_{k}=1,0$ if ball $k$ is in urn I, II, respectively, one may identify the states with the vertices of the N-cube. $a$ $b$ Let and denote real numbers such that $a,$ $b>-1$ or $a,$ $b<-N$ , and consider the following homogeneous Morkov chain with state space $\{0,1, \ldots , N\}$ and transition probabilities $(1- \frac{i}{N})(\frac{a+1+i}{N+a+b+2})l\text{ノ}$, if $j=i+1$ , $\frac{i}{N}(\frac{N+b+1-i}{N+a+b+2})\nu$ , if $j=i-1$ , $p_{ij}=\{$ (1.2) $1-(1- \frac{i}{N})(\frac{a+1+i}{N+a+b+2})$ $l \text{ノ}-\frac{i}{N}(\frac{N+b+1-i}{N+a+b+2})l^{\text{ノ}}$, if $j=i$ , $0$ , otherwise, $\nu>0$ $0=^{pij=^{1}}<<$ $N$ where is arbitrary such that for all $i,j=0,1,$ $\ldots,$ . Remark 1.2. The two-parameter Ehrenfest urn model (1.1) is obtained by putting $a=su/(s+t),$ $b=tu/(s+t)$ and $\nu=s+t$ and taking the limit as $uarrow\infty$ . For the model (1.2), Dette [4] obtains the more general integral representations for the $n$-step transition probabilities and the (unique) stationary probability distribution with respect to the spectral measure of random walk and shows the following remark. Remark 1.3. (i) The $n$ -step transition probabilities $p_{ik}^{(n)}$ in the model (1.1) are given by $p_{ik}^{(n)}=( \frac{p}{q})^{k}\sum_{=x\mathrm{o}}^{N}p^{x}q^{N\prime}-x_{I\iota(X}i,p,$ $N)I\mathrm{i}_{k}(')x,p,$$N(1- \frac{s+t}{N}X)^{n}$ , where $p=s/(s+t)$ and $q=t/(s+t)$ . Here $\{K_{n}(x,p, N) : n=0,1, \ldots, N\}$ is the family of Krawtchouk polynomials, which will appear in Remark 1.4 below. This result is also found by Krafft and Schaefer [9]. (ii) The (unique) stationary probability distribution $\pi=(\pi_{0}, \ldots , \pi_{N})$ in the model (1.2) is given by $\pi_{j}=\frac{\beta(a+1+j,N+b+1-j)}{\beta(a+1,b+1)}$ , where $\beta(x, y)$ denotes the beta function, that is, $\beta(x, y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}=\int_{0}^{1}u^{x-1}(1-u)^{y-}1du$ for $x,$ $y>0$ . 81 For the model (1.1) with $s=t=1$ , Kac [6] finds the same form with (i) of Remark 1.3 and shows Newton’s law of cooling that concerns the estimate of the excess of the numbers of balls in urn I over the mean of the stationary probability distribution. Our purpose is the following: $\bullet$ For the model (1.2), we will show the same result with Newton’s law of cooling [Theorem 2.1]. $\bullet$ For the model (1.1), we will give the continuous-time formulation and obtain the law of transition probabilities and first-passage times in both the ‘reduced’ and the ‘full’ descriptions, using the method of Bingham [3] [Theorems 2.2-2.3]. Remark 1.4. With $0<p=1-q<1$ and integer $N>=0$ the Krawtchouk polynomials are defined by the hypergeometric series: $K_{n}(x,p, N)$ $=$ $2F1(-n, -x;-N;1/p)$ $=$ $\sum_{\nu=0}^{n}(-1)^{\nu}\frac{(\begin{array}{l}n\nu\end{array})(\begin{array}{l}x\nu\end{array})}{(\begin{array}{l}N\nu\end{array})}(\frac{1}{p})^{\nu}$ $N$ , $n=0,1,$ $\ldots,$ . When the meaning is clear, we write simply $K_{n}(x)$ instead of $K_{n}(x,p, N)$ . From Karlin $\mathrm{M}_{\mathrm{C}}\mathrm{c}\mathrm{r}\mathrm{e}\mathrm{g}_{\mathrm{o}\mathrm{r}}[7]$ and we cite the properties for $I_{\mathrm{L}_{n}}'(X)$ , which will be used in after sections. (i) $\sum_{n=0}^{N}I\mathrm{i}_{n}'(X)_{Z}n=(1+z)N-x(1-\frac{q}{p}Z)^{x}$ $N$ , $x=0,1,$ $\ldots,$ . (ii) $I\mathrm{i}_{n}^{r}(0)=1$ , $I\mathrm{i}_{0}^{r}(x)=1$ . $I\mathrm{i}_{n}^{\Gamma}(x, p, N)=I\mathrm{i}_{x}^{r}(n,p, N)$ $n,$ $N$ (iii) , $x=0,1,$ $\ldots,$ . (iv) $I \mathrm{i}_{n}^{r}(N-x, p, N)=(-1)^{n}(\frac{q}{p})^{n}I\iota_{n}'(x, q, N)$ . 2. Theorems We first concern ourselves with evaluation of the excess of the number of balls in urn I over the value averaged by the stationary probability distribution for the model (1.2). Theorem 2.1. Let $\{X_{n} : n=0,1, \ldots\}$ be the Markov chain with transition probabilities (1.2) with $a>-1$ and $b>-1$ . Let $\pi=(\pi_{0}, \ldots , \pi_{N})$ be the stationary probabilitity distribution given by (ii) of Remark 1.3, and denote $by<\pi>the$ mean of $\pi$ , that is, $< \pi>=\sum_{j0}^{N}=j\pi_{j}$ . Set $Y_{n}=X_{n}-<\pi>and$ put $e_{n}=E_{i}(Y_{n})$ , that is, the expected value of $Y_{n}$ given $X_{0}=i$ . Then we have the following: (i) $< \pi>=N(\frac{a+1}{a+b+2})$ . 82 (ii) $e_{n}=(i-< \pi>)[1-\nu(\frac{a+b+2}{N(N+a+b+2)})]^{n}$ . $T$ Remark 2.1. Suppose that the frequency of transitions is $\tau$ per second. Then in time there are $n=T\tau$ transitions. Write $c=- \log[1-\nu(\frac{a+b+2}{N(N+a+b+2)})]^{\mathcal{T}}$ Then (ii) of the above theorem yields $e_{n}=(i-<\pi>)\exp[-cT]$ . (2.1) $b$ Let $a,$ and $\nu$ be taken as in Remark 1.2. Then $< \pi>arrow N(\frac{s}{s+t})$ and l–v $( \frac{a+b+2}{N(N+a+b+2)})arrow 1-\frac{s+t}{N}$ in the limit as $uarrow\infty$ , which shows $e_{n}= \{i-N(\frac{s}{s+t})\}\exp[-C\tau]$ with $c=- \log[1-\frac{s+t}{N}]^{\tau}$ (2.2) Newton’s law of cooling in Kac [6] is the special case of (2.2) where $s=t=1$ .
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